(a) The mathematical proof of the statement stating that using L'Hopital Rule, f ( x ) = e − x − e − a x x , though not defined at x = 0, can be made continuous by assigning the value f ( 0 ) = a − 1

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Answer 1

Question

Chapter 15.1, Problem 53E

To determine

(a)

The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$

Expert Solution

Answer to Problem 53E

Solution: The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$ is derived.

Explanation of Solution

Explanation:

Given: $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ for a>0

Calculation:

Step 1: The function $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , $f (0) = a - 1$ is continuous if $\lim_{x \to 0} f (x) = f (0) = a - 1$ .

Step 2: We verify this limit using L'Hopital Rule:

$\lim_{x \to 0} \frac{e^{- x} - e^{- a x}}{x} = \lim_{x \to 0} \frac{- e^{- x} + a e^{- a x}}{1} = - 1 + a = a - 1$

Therefore, f is continuous.

Conclusion: The statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$ is mathematically proved.

To determine

(b)

The mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem

Expert Solution

Answer to Problem 53E

Solution: Both the mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem are derived.

Explanation of Solution

Given: $x > 1$ , $I (a) = \int_{0}^{\infty} \frac{e^{- x} - e^{- a x}}{x} d x$ , $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , a > 0

Calculation:

Step 1: We now show that the following integral converges:

$I (a) = \int_{0}^{\infty} \frac{e^{- x} - e^{- a x}}{x} d x$ (a>0)

Since, $e^{- x} - e^{- a x} < e^{- x} + e^{- a x}$ then $\frac{e^{- x} - e^{- a x}}{x} < \frac{e^{- x} + e^{- a x}}{x}$ for x > 0

If x > 1 we have,

$\frac{e^{- x} - e^{- a x}}{x} < \frac{e^{- x} + e^{- a x}}{x} < e^{- x} + e^{- a x}$

That is for x > 1

$f (x) < e^{- x} + e^{- a x}$

Step 2: Also, since $e^{- a x} - e^{- x} < e^{- a x} + e^{- x}$ we have for x > 1

$\frac{e^{- a x} - e^{- x}}{x} < \frac{e^{- a x} - e^{- x}}{x} < e^{- a x} + e^{- x}$

Thus, we get

$- f (x) < e^{- x} + e^{- a x}$

Step 3: Hence, from Step 1 and Step 2, we get

$0 \leq | f (x) | \leq e^{- x} + e^{- a x}$

Step 4: We now show that the integral of the right hand side converges:

$\begin{array}{l} \int_{0}^{\infty} (e^{- x} + e^{- a x}) d x = \lim_{R \to \infty} \int_{0}^{R} (e^{- x} + e^{- a x}) d x = \lim_{R \to \infty} (- e^{- x} - {\frac{e^{- a x}}{a} |}_{x = 0}^{R}) \\ = \lim_{R \to \infty} (e^{- R} - \frac{e^{- a R}}{a} + e^{0} + \frac{e^{0}}{a}) = \lim_{R \to \infty} (e^{- R} - \frac{e^{- a R}}{a} + 1 + \frac{1}{a}) = 1 + \frac{1}{a} \end{array}$

Since the integral converges, we conclude from Step 3 and the Comparison Test for Improper Integral that

$\int_{0}^{\infty} \frac{(e^{- x} + e^{- a x})}{x} d x$ also converges for a > 0.

Conclusion: Both the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the statement stating that I(a) converges by applying the Comparison Theorem are mathematically proved.

To determine

(c)

The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$

Expert Solution

Answer to Problem 53E

Solution: The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$ is derived

Explanation of Solution

Given: $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$ , a>0

Calculation:

Step 1: We compute the inner integral with respect to y:

$\int_{1}^{a} e^{- x y} d y = - \frac{1}{x} {e^{- x y} |}_{y = 1}^{a} = - \frac{1}{x} (e^{- x a} - e^{- x .1}) = \frac{e^{- x} - e^{- a x}}{x}$

Step 2: Hence,

$I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x = \int_{0}^{\infty} (\int_{1}^{a} e^{- x y} d y) d x = \int_{0}^{\infty} (\frac{e^{- x} - e^{- a x}}{x}) d x = I (a)$

Conclusion: The equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$ is mathematically proved.

To determine

(d)

By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

Expert Solution

Answer to Problem 53E

Solution: By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is derived.

Explanation of Solution

Given: $I (a) = \int_{0}^{T} \int_{1}^{a} e^{- x y} d y d x$ , a>0

Calculation:

Step 1: By the definition of the improper integral,

$I (a) = \lim_{T \to \infty} \int_{0}^{T} \int_{1}^{a} e^{- x y} d y d x$

Step 2: We compute the double integral. Using Fubini's Theorem we may compute the iterated integral using the reversed order of integration. That is,

$\begin{array}{l} \int_{0}^{T} \int_{1}^{a} e^{- x y} d y d x = \int_{1}^{a} \int_{0}^{T} e^{- x y} d x d y = \int_{1}^{a} (\int_{0}^{T} e^{- x y} d x) d y = \int_{1}^{a} ({- \frac{1}{y} e^{- x y} |}_{x = 0}^{T}) d y \\ = \int_{1}^{a} (- \frac{1}{y} (e^{- T y} - e^{- 0. y})) d y = \int_{1}^{a} \frac{1 - e^{- T y}}{y} d y = \int_{1}^{a} \frac{d y}{y} - \int_{1}^{a} \frac{e^{- T y}}{y} d y \\ = {\ln y |}_{1}^{a} - \int_{1}^{a} \frac{e^{- T y}}{y} d y = \ln a - \ln 1 - \int_{1}^{a} \frac{e^{- T y}}{y} d y = \ln a - \int_{1}^{a} \frac{e^{- T y}}{y} d y \end{array}$

Combining with Step 1, we get,

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

Conclusion: By interchanging the order of integration, the equation $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is mathematically proved.

To determine

(e)

The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem.

Expert Solution

Answer to Problem 53E

Solution: The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem is derived.

Explanation of Solution

Given: $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ , a>0

Calculation:

Step 1: We consider the following possible cases:

Case 1: $a \geq 1$ then in the interval of integration $y \geq 1$ . As $T \to \infty$ , we may assume that T > 0

Thus,

$\frac{e^{- T y}}{y} \leq \frac{e^{- T .1}}{1} = e^{- T}$

Hence,

$0 \leq \int_{1}^{a} \frac{e^{- T y}}{y} d y \leq \int_{1}^{a} e^{- T} d y = e^{- T} (a - 1)$

By the limit $\lim_{T \to \infty} e^{- T} (a - 1) = 0$ and the Squeeze Theorem, we conclude that,

$\lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y = 0$

Case 2: $0 < a < 1.$

Then,

$\int_{1}^{a} \frac{e^{- T y}}{y} d y = - \int_{a}^{1} \frac{e^{- T y}}{y} d y$

and in the interval of integration $a \leq y \leq 1$ , therefore

$\frac{e^{- T y}}{y} \leq \frac{e^{- T a}}{a}$ (the function $\frac{e^{- T y}}{y}$ is decreasing).

Hence,

$0 \leq \int_{a}^{1} \frac{e^{- T y}}{y} d y \leq \int_{a}^{1} \frac{e^{- T a}}{a} d y = \frac{1 - a}{a} e^{- T a}$

By the limit $\lim_{T \to \infty} \frac{1 - a}{a} e^{- T a} = 0$ and the Squeeze Theorem, we conclude that

$\lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} = - \lim_{T \to \infty} \int_{a}^{1} \frac{e^{- T y}}{y} = 0$

We thus showed that for all a > 0, $\lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} = 0$

Conclusion: The statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem is mathematically proved.

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Answer 2

Question

Chapter 15.1, Problem 53E

To determine

(a)

The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$

Expert Solution

Answer to Problem 53E

Solution: The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$ is derived.

Explanation of Solution

Explanation:

Given: $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ for a>0

Calculation:

Step 1: The function $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , $f (0) = a - 1$ is continuous if $\lim_{x \to 0} f (x) = f (0) = a - 1$ .

Step 2: We verify this limit using L'Hopital Rule:

$\lim_{x \to 0} \frac{e^{- x} - e^{- a x}}{x} = \lim_{x \to 0} \frac{- e^{- x} + a e^{- a x}}{1} = - 1 + a = a - 1$

Therefore, f is continuous.

Conclusion: The statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$ is mathematically proved.

To determine

(b)

The mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem

Expert Solution

Answer to Problem 53E

Solution: Both the mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem are derived.

Explanation of Solution

Given: $x > 1$ , $I (a) = \int_{0}^{\infty} \frac{e^{- x} - e^{- a x}}{x} d x$ , $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , a > 0

Calculation:

Step 1: We now show that the following integral converges:

$I (a) = \int_{0}^{\infty} \frac{e^{- x} - e^{- a x}}{x} d x$ (a>0)

Since, $e^{- x} - e^{- a x} < e^{- x} + e^{- a x}$ then $\frac{e^{- x} - e^{- a x}}{x} < \frac{e^{- x} + e^{- a x}}{x}$ for x > 0

If x > 1 we have,

$\frac{e^{- x} - e^{- a x}}{x} < \frac{e^{- x} + e^{- a x}}{x} < e^{- x} + e^{- a x}$

That is for x > 1

$f (x) < e^{- x} + e^{- a x}$

Step 2: Also, since $e^{- a x} - e^{- x} < e^{- a x} + e^{- x}$ we have for x > 1

$\frac{e^{- a x} - e^{- x}}{x} < \frac{e^{- a x} - e^{- x}}{x} < e^{- a x} + e^{- x}$

Thus, we get

$- f (x) < e^{- x} + e^{- a x}$

Step 3: Hence, from Step 1 and Step 2, we get

$0 \leq | f (x) | \leq e^{- x} + e^{- a x}$

Step 4: We now show that the integral of the right hand side converges:

$\begin{array}{l} \int_{0}^{\infty} (e^{- x} + e^{- a x}) d x = \lim_{R \to \infty} \int_{0}^{R} (e^{- x} + e^{- a x}) d x = \lim_{R \to \infty} (- e^{- x} - {\frac{e^{- a x}}{a} |}_{x = 0}^{R}) \\ = \lim_{R \to \infty} (e^{- R} - \frac{e^{- a R}}{a} + e^{0} + \frac{e^{0}}{a}) = \lim_{R \to \infty} (e^{- R} - \frac{e^{- a R}}{a} + 1 + \frac{1}{a}) = 1 + \frac{1}{a} \end{array}$

Since the integral converges, we conclude from Step 3 and the Comparison Test for Improper Integral that

$\int_{0}^{\infty} \frac{(e^{- x} + e^{- a x})}{x} d x$ also converges for a > 0.

Conclusion: Both the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the statement stating that I(a) converges by applying the Comparison Theorem are mathematically proved.

To determine

(c)

The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$

Expert Solution

Answer to Problem 53E

Solution: The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$ is derived

Explanation of Solution

Given: $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$ , a>0

Calculation:

Step 1: We compute the inner integral with respect to y:

$\int_{1}^{a} e^{- x y} d y = - \frac{1}{x} {e^{- x y} |}_{y = 1}^{a} = - \frac{1}{x} (e^{- x a} - e^{- x .1}) = \frac{e^{- x} - e^{- a x}}{x}$

Step 2: Hence,

$I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x = \int_{0}^{\infty} (\int_{1}^{a} e^{- x y} d y) d x = \int_{0}^{\infty} (\frac{e^{- x} - e^{- a x}}{x}) d x = I (a)$

Conclusion: The equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$ is mathematically proved.

To determine

(d)

By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

Expert Solution

Answer to Problem 53E

Solution: By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is derived.

Explanation of Solution

Given: $I (a) = \int_{0}^{T} \int_{1}^{a} e^{- x y} d y d x$ , a>0

Calculation:

Step 1: By the definition of the improper integral,

$I (a) = \lim_{T \to \infty} \int_{0}^{T} \int_{1}^{a} e^{- x y} d y d x$

Step 2: We compute the double integral. Using Fubini's Theorem we may compute the iterated integral using the reversed order of integration. That is,

$\begin{array}{l} \int_{0}^{T} \int_{1}^{a} e^{- x y} d y d x = \int_{1}^{a} \int_{0}^{T} e^{- x y} d x d y = \int_{1}^{a} (\int_{0}^{T} e^{- x y} d x) d y = \int_{1}^{a} ({- \frac{1}{y} e^{- x y} |}_{x = 0}^{T}) d y \\ = \int_{1}^{a} (- \frac{1}{y} (e^{- T y} - e^{- 0. y})) d y = \int_{1}^{a} \frac{1 - e^{- T y}}{y} d y = \int_{1}^{a} \frac{d y}{y} - \int_{1}^{a} \frac{e^{- T y}}{y} d y \\ = {\ln y |}_{1}^{a} - \int_{1}^{a} \frac{e^{- T y}}{y} d y = \ln a - \ln 1 - \int_{1}^{a} \frac{e^{- T y}}{y} d y = \ln a - \int_{1}^{a} \frac{e^{- T y}}{y} d y \end{array}$

Combining with Step 1, we get,

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

Conclusion: By interchanging the order of integration, the equation $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is mathematically proved.

To determine

(e)

The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem.

Expert Solution

Answer to Problem 53E

Solution: The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem is derived.

Explanation of Solution

Given: $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ , a>0

Calculation:

Step 1: We consider the following possible cases:

Case 1: $a \geq 1$ then in the interval of integration $y \geq 1$ . As $T \to \infty$ , we may assume that T > 0

Thus,

$\frac{e^{- T y}}{y} \leq \frac{e^{- T .1}}{1} = e^{- T}$

Hence,

$0 \leq \int_{1}^{a} \frac{e^{- T y}}{y} d y \leq \int_{1}^{a} e^{- T} d y = e^{- T} (a - 1)$

By the limit $\lim_{T \to \infty} e^{- T} (a - 1) = 0$ and the Squeeze Theorem, we conclude that,

$\lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y = 0$

Case 2: $0 < a < 1.$

Then,

$\int_{1}^{a} \frac{e^{- T y}}{y} d y = - \int_{a}^{1} \frac{e^{- T y}}{y} d y$

and in the interval of integration $a \leq y \leq 1$ , therefore

$\frac{e^{- T y}}{y} \leq \frac{e^{- T a}}{a}$ (the function $\frac{e^{- T y}}{y}$ is decreasing).

Hence,

$0 \leq \int_{a}^{1} \frac{e^{- T y}}{y} d y \leq \int_{a}^{1} \frac{e^{- T a}}{a} d y = \frac{1 - a}{a} e^{- T a}$

By the limit $\lim_{T \to \infty} \frac{1 - a}{a} e^{- T a} = 0$ and the Squeeze Theorem, we conclude that

$\lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} = - \lim_{T \to \infty} \int_{a}^{1} \frac{e^{- T y}}{y} = 0$

We thus showed that for all a > 0, $\lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} = 0$

Conclusion: The statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem is mathematically proved.

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Answer 3

Question

Chapter 15.1, Problem 53E

To determine

(a)

The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$

Expert Solution

Answer to Problem 53E

Solution: The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$ is derived.

Explanation of Solution

Explanation:

Given: $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ for a>0

Calculation:

Step 1: The function $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , $f (0) = a - 1$ is continuous if $\lim_{x \to 0} f (x) = f (0) = a - 1$ .

Step 2: We verify this limit using L'Hopital Rule:

$\lim_{x \to 0} \frac{e^{- x} - e^{- a x}}{x} = \lim_{x \to 0} \frac{- e^{- x} + a e^{- a x}}{1} = - 1 + a = a - 1$

Therefore, f is continuous.

Conclusion: The statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$ is mathematically proved.

To determine

(b)

The mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem

Expert Solution

Answer to Problem 53E

Solution: Both the mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem are derived.

Explanation of Solution

Given: $x > 1$ , $I (a) = \int_{0}^{\infty} \frac{e^{- x} - e^{- a x}}{x} d x$ , $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , a > 0

Calculation:

Step 1: We now show that the following integral converges:

$I (a) = \int_{0}^{\infty} \frac{e^{- x} - e^{- a x}}{x} d x$ (a>0)

Since, $e^{- x} - e^{- a x} < e^{- x} + e^{- a x}$ then $\frac{e^{- x} - e^{- a x}}{x} < \frac{e^{- x} + e^{- a x}}{x}$ for x > 0

If x > 1 we have,

$\frac{e^{- x} - e^{- a x}}{x} < \frac{e^{- x} + e^{- a x}}{x} < e^{- x} + e^{- a x}$

That is for x > 1

$f (x) < e^{- x} + e^{- a x}$

Step 2: Also, since $e^{- a x} - e^{- x} < e^{- a x} + e^{- x}$ we have for x > 1

$\frac{e^{- a x} - e^{- x}}{x} < \frac{e^{- a x} - e^{- x}}{x} < e^{- a x} + e^{- x}$

Thus, we get

$- f (x) < e^{- x} + e^{- a x}$

Step 3: Hence, from Step 1 and Step 2, we get

$0 \leq | f (x) | \leq e^{- x} + e^{- a x}$

Step 4: We now show that the integral of the right hand side converges:

$\begin{array}{l} \int_{0}^{\infty} (e^{- x} + e^{- a x}) d x = \lim_{R \to \infty} \int_{0}^{R} (e^{- x} + e^{- a x}) d x = \lim_{R \to \infty} (- e^{- x} - {\frac{e^{- a x}}{a} |}_{x = 0}^{R}) \\ = \lim_{R \to \infty} (e^{- R} - \frac{e^{- a R}}{a} + e^{0} + \frac{e^{0}}{a}) = \lim_{R \to \infty} (e^{- R} - \frac{e^{- a R}}{a} + 1 + \frac{1}{a}) = 1 + \frac{1}{a} \end{array}$

Since the integral converges, we conclude from Step 3 and the Comparison Test for Improper Integral that

$\int_{0}^{\infty} \frac{(e^{- x} + e^{- a x})}{x} d x$ also converges for a > 0.

Conclusion: Both the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the statement stating that I(a) converges by applying the Comparison Theorem are mathematically proved.

To determine

(c)

The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$

Expert Solution

Answer to Problem 53E

Solution: The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$ is derived

Explanation of Solution

Given: $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$ , a>0

Calculation:

Step 1: We compute the inner integral with respect to y:

$\int_{1}^{a} e^{- x y} d y = - \frac{1}{x} {e^{- x y} |}_{y = 1}^{a} = - \frac{1}{x} (e^{- x a} - e^{- x .1}) = \frac{e^{- x} - e^{- a x}}{x}$

Step 2: Hence,

$I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x = \int_{0}^{\infty} (\int_{1}^{a} e^{- x y} d y) d x = \int_{0}^{\infty} (\frac{e^{- x} - e^{- a x}}{x}) d x = I (a)$

Conclusion: The equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$ is mathematically proved.

To determine

(d)

By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

Expert Solution

Answer to Problem 53E

Solution: By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is derived.

Explanation of Solution

Given: $I (a) = \int_{0}^{T} \int_{1}^{a} e^{- x y} d y d x$ , a>0

Calculation:

Step 1: By the definition of the improper integral,

$I (a) = \lim_{T \to \infty} \int_{0}^{T} \int_{1}^{a} e^{- x y} d y d x$

Step 2: We compute the double integral. Using Fubini's Theorem we may compute the iterated integral using the reversed order of integration. That is,

$\begin{array}{l} \int_{0}^{T} \int_{1}^{a} e^{- x y} d y d x = \int_{1}^{a} \int_{0}^{T} e^{- x y} d x d y = \int_{1}^{a} (\int_{0}^{T} e^{- x y} d x) d y = \int_{1}^{a} ({- \frac{1}{y} e^{- x y} |}_{x = 0}^{T}) d y \\ = \int_{1}^{a} (- \frac{1}{y} (e^{- T y} - e^{- 0. y})) d y = \int_{1}^{a} \frac{1 - e^{- T y}}{y} d y = \int_{1}^{a} \frac{d y}{y} - \int_{1}^{a} \frac{e^{- T y}}{y} d y \\ = {\ln y |}_{1}^{a} - \int_{1}^{a} \frac{e^{- T y}}{y} d y = \ln a - \ln 1 - \int_{1}^{a} \frac{e^{- T y}}{y} d y = \ln a - \int_{1}^{a} \frac{e^{- T y}}{y} d y \end{array}$

Combining with Step 1, we get,

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

Conclusion: By interchanging the order of integration, the equation $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is mathematically proved.

To determine

(e)

The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem.

Expert Solution

Answer to Problem 53E

Solution: The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem is derived.

Explanation of Solution

Given: $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ , a>0

Calculation:

Step 1: We consider the following possible cases:

Case 1: $a \geq 1$ then in the interval of integration $y \geq 1$ . As $T \to \infty$ , we may assume that T > 0

Thus,

$\frac{e^{- T y}}{y} \leq \frac{e^{- T .1}}{1} = e^{- T}$

Hence,

$0 \leq \int_{1}^{a} \frac{e^{- T y}}{y} d y \leq \int_{1}^{a} e^{- T} d y = e^{- T} (a - 1)$

By the limit $\lim_{T \to \infty} e^{- T} (a - 1) = 0$ and the Squeeze Theorem, we conclude that,

$\lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y = 0$

Case 2: $0 < a < 1.$

Then,

$\int_{1}^{a} \frac{e^{- T y}}{y} d y = - \int_{a}^{1} \frac{e^{- T y}}{y} d y$

and in the interval of integration $a \leq y \leq 1$ , therefore

$\frac{e^{- T y}}{y} \leq \frac{e^{- T a}}{a}$ (the function $\frac{e^{- T y}}{y}$ is decreasing).

Hence,

$0 \leq \int_{a}^{1} \frac{e^{- T y}}{y} d y \leq \int_{a}^{1} \frac{e^{- T a}}{a} d y = \frac{1 - a}{a} e^{- T a}$

By the limit $\lim_{T \to \infty} \frac{1 - a}{a} e^{- T a} = 0$ and the Squeeze Theorem, we conclude that

$\lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} = - \lim_{T \to \infty} \int_{a}^{1} \frac{e^{- T y}}{y} = 0$

We thus showed that for all a > 0, $\lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} = 0$

Conclusion: The statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem is mathematically proved.

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Answer 4

Question

Chapter 15.1, Problem 53E

To determine

(a)

The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$

Expert Solution

Answer to Problem 53E

Solution: The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$ is derived.

Explanation of Solution

Explanation:

Given: $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ for a>0

Calculation:

Step 1: The function $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , $f (0) = a - 1$ is continuous if $\lim_{x \to 0} f (x) = f (0) = a - 1$ .

Step 2: We verify this limit using L'Hopital Rule:

$\lim_{x \to 0} \frac{e^{- x} - e^{- a x}}{x} = \lim_{x \to 0} \frac{- e^{- x} + a e^{- a x}}{1} = - 1 + a = a - 1$

Therefore, f is continuous.

Conclusion: The statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$ is mathematically proved.

To determine

(b)

The mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem

Expert Solution

Answer to Problem 53E

Solution: Both the mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem are derived.

Explanation of Solution

Given: $x > 1$ , $I (a) = \int_{0}^{\infty} \frac{e^{- x} - e^{- a x}}{x} d x$ , $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , a > 0

Calculation:

Step 1: We now show that the following integral converges:

$I (a) = \int_{0}^{\infty} \frac{e^{- x} - e^{- a x}}{x} d x$ (a>0)

Since, $e^{- x} - e^{- a x} < e^{- x} + e^{- a x}$ then $\frac{e^{- x} - e^{- a x}}{x} < \frac{e^{- x} + e^{- a x}}{x}$ for x > 0

If x > 1 we have,

$\frac{e^{- x} - e^{- a x}}{x} < \frac{e^{- x} + e^{- a x}}{x} < e^{- x} + e^{- a x}$

That is for x > 1

$f (x) < e^{- x} + e^{- a x}$

Step 2: Also, since $e^{- a x} - e^{- x} < e^{- a x} + e^{- x}$ we have for x > 1

$\frac{e^{- a x} - e^{- x}}{x} < \frac{e^{- a x} - e^{- x}}{x} < e^{- a x} + e^{- x}$

Thus, we get

$- f (x) < e^{- x} + e^{- a x}$

Step 3: Hence, from Step 1 and Step 2, we get

$0 \leq | f (x) | \leq e^{- x} + e^{- a x}$

Step 4: We now show that the integral of the right hand side converges:

$\begin{array}{l} \int_{0}^{\infty} (e^{- x} + e^{- a x}) d x = \lim_{R \to \infty} \int_{0}^{R} (e^{- x} + e^{- a x}) d x = \lim_{R \to \infty} (- e^{- x} - {\frac{e^{- a x}}{a} |}_{x = 0}^{R}) \\ = \lim_{R \to \infty} (e^{- R} - \frac{e^{- a R}}{a} + e^{0} + \frac{e^{0}}{a}) = \lim_{R \to \infty} (e^{- R} - \frac{e^{- a R}}{a} + 1 + \frac{1}{a}) = 1 + \frac{1}{a} \end{array}$

Since the integral converges, we conclude from Step 3 and the Comparison Test for Improper Integral that

$\int_{0}^{\infty} \frac{(e^{- x} + e^{- a x})}{x} d x$ also converges for a > 0.

Conclusion: Both the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the statement stating that I(a) converges by applying the Comparison Theorem are mathematically proved.

To determine

(c)

The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$

Expert Solution

Answer to Problem 53E

Solution: The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$ is derived

Explanation of Solution

Given: $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$ , a>0

Calculation:

Step 1: We compute the inner integral with respect to y:

$\int_{1}^{a} e^{- x y} d y = - \frac{1}{x} {e^{- x y} |}_{y = 1}^{a} = - \frac{1}{x} (e^{- x a} - e^{- x .1}) = \frac{e^{- x} - e^{- a x}}{x}$

Step 2: Hence,

$I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x = \int_{0}^{\infty} (\int_{1}^{a} e^{- x y} d y) d x = \int_{0}^{\infty} (\frac{e^{- x} - e^{- a x}}{x}) d x = I (a)$

Conclusion: The equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$ is mathematically proved.

To determine

(d)

By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

Expert Solution

Answer to Problem 53E

Solution: By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is derived.

Explanation of Solution

Given: $I (a) = \int_{0}^{T} \int_{1}^{a} e^{- x y} d y d x$ , a>0

Calculation:

Step 1: By the definition of the improper integral,

$I (a) = \lim_{T \to \infty} \int_{0}^{T} \int_{1}^{a} e^{- x y} d y d x$

Step 2: We compute the double integral. Using Fubini's Theorem we may compute the iterated integral using the reversed order of integration. That is,

$\begin{array}{l} \int_{0}^{T} \int_{1}^{a} e^{- x y} d y d x = \int_{1}^{a} \int_{0}^{T} e^{- x y} d x d y = \int_{1}^{a} (\int_{0}^{T} e^{- x y} d x) d y = \int_{1}^{a} ({- \frac{1}{y} e^{- x y} |}_{x = 0}^{T}) d y \\ = \int_{1}^{a} (- \frac{1}{y} (e^{- T y} - e^{- 0. y})) d y = \int_{1}^{a} \frac{1 - e^{- T y}}{y} d y = \int_{1}^{a} \frac{d y}{y} - \int_{1}^{a} \frac{e^{- T y}}{y} d y \\ = {\ln y |}_{1}^{a} - \int_{1}^{a} \frac{e^{- T y}}{y} d y = \ln a - \ln 1 - \int_{1}^{a} \frac{e^{- T y}}{y} d y = \ln a - \int_{1}^{a} \frac{e^{- T y}}{y} d y \end{array}$

Combining with Step 1, we get,

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

Conclusion: By interchanging the order of integration, the equation $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is mathematically proved.

To determine

(e)

The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem.

Expert Solution

Answer to Problem 53E

Solution: The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem is derived.

Explanation of Solution

Given: $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ , a>0

Calculation:

Step 1: We consider the following possible cases:

Case 1: $a \geq 1$ then in the interval of integration $y \geq 1$ . As $T \to \infty$ , we may assume that T > 0

Thus,

$\frac{e^{- T y}}{y} \leq \frac{e^{- T .1}}{1} = e^{- T}$

Hence,

$0 \leq \int_{1}^{a} \frac{e^{- T y}}{y} d y \leq \int_{1}^{a} e^{- T} d y = e^{- T} (a - 1)$

By the limit $\lim_{T \to \infty} e^{- T} (a - 1) = 0$ and the Squeeze Theorem, we conclude that,

$\lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y = 0$

Case 2: $0 < a < 1.$

Then,

$\int_{1}^{a} \frac{e^{- T y}}{y} d y = - \int_{a}^{1} \frac{e^{- T y}}{y} d y$

and in the interval of integration $a \leq y \leq 1$ , therefore

$\frac{e^{- T y}}{y} \leq \frac{e^{- T a}}{a}$ (the function $\frac{e^{- T y}}{y}$ is decreasing).

Hence,

$0 \leq \int_{a}^{1} \frac{e^{- T y}}{y} d y \leq \int_{a}^{1} \frac{e^{- T a}}{a} d y = \frac{1 - a}{a} e^{- T a}$

By the limit $\lim_{T \to \infty} \frac{1 - a}{a} e^{- T a} = 0$ and the Squeeze Theorem, we conclude that

$\lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} = - \lim_{T \to \infty} \int_{a}^{1} \frac{e^{- T y}}{y} = 0$

We thus showed that for all a > 0, $\lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} = 0$

Conclusion: The statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem is mathematically proved.

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Answer 5

Chapter 15.1, Problem 53E

To determine

(a)

The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$

Expert Solution

Answer to Problem 53E

Solution: The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$ is derived.

Explanation of Solution

Explanation:

Given: $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ for a>0

Calculation:

Step 1: The function $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , $f (0) = a - 1$ is continuous if $\lim_{x \to 0} f (x) = f (0) = a - 1$ .

Step 2: We verify this limit using L'Hopital Rule:

$\lim_{x \to 0} \frac{e^{- x} - e^{- a x}}{x} = \lim_{x \to 0} \frac{- e^{- x} + a e^{- a x}}{1} = - 1 + a = a - 1$

Therefore, f is continuous.

Conclusion: The statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$ is mathematically proved.

To determine

(b)

The mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem

Expert Solution

Answer to Problem 53E

Solution: Both the mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem are derived.

Explanation of Solution

Given: $x > 1$ , $I (a) = \int_{0}^{\infty} \frac{e^{- x} - e^{- a x}}{x} d x$ , $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , a > 0

Calculation:

Step 1: We now show that the following integral converges:

$I (a) = \int_{0}^{\infty} \frac{e^{- x} - e^{- a x}}{x} d x$ (a>0)

Since, $e^{- x} - e^{- a x} < e^{- x} + e^{- a x}$ then $\frac{e^{- x} - e^{- a x}}{x} < \frac{e^{- x} + e^{- a x}}{x}$ for x > 0

If x > 1 we have,

$\frac{e^{- x} - e^{- a x}}{x} < \frac{e^{- x} + e^{- a x}}{x} < e^{- x} + e^{- a x}$

That is for x > 1

$f (x) < e^{- x} + e^{- a x}$

Step 2: Also, since $e^{- a x} - e^{- x} < e^{- a x} + e^{- x}$ we have for x > 1

$\frac{e^{- a x} - e^{- x}}{x} < \frac{e^{- a x} - e^{- x}}{x} < e^{- a x} + e^{- x}$

Thus, we get

$- f (x) < e^{- x} + e^{- a x}$

Step 3: Hence, from Step 1 and Step 2, we get

$0 \leq | f (x) | \leq e^{- x} + e^{- a x}$

Step 4: We now show that the integral of the right hand side converges:

$\begin{array}{l} \int_{0}^{\infty} (e^{- x} + e^{- a x}) d x = \lim_{R \to \infty} \int_{0}^{R} (e^{- x} + e^{- a x}) d x = \lim_{R \to \infty} (- e^{- x} - {\frac{e^{- a x}}{a} |}_{x = 0}^{R}) \\ = \lim_{R \to \infty} (e^{- R} - \frac{e^{- a R}}{a} + e^{0} + \frac{e^{0}}{a}) = \lim_{R \to \infty} (e^{- R} - \frac{e^{- a R}}{a} + 1 + \frac{1}{a}) = 1 + \frac{1}{a} \end{array}$

Since the integral converges, we conclude from Step 3 and the Comparison Test for Improper Integral that

$\int_{0}^{\infty} \frac{(e^{- x} + e^{- a x})}{x} d x$ also converges for a > 0.

Conclusion: Both the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the statement stating that I(a) converges by applying the Comparison Theorem are mathematically proved.

To determine

(c)

The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$

Expert Solution

Answer to Problem 53E

Solution: The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$ is derived

Explanation of Solution

Given: $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$ , a>0

Calculation:

Step 1: We compute the inner integral with respect to y:

$\int_{1}^{a} e^{- x y} d y = - \frac{1}{x} {e^{- x y} |}_{y = 1}^{a} = - \frac{1}{x} (e^{- x a} - e^{- x .1}) = \frac{e^{- x} - e^{- a x}}{x}$

Step 2: Hence,

$I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x = \int_{0}^{\infty} (\int_{1}^{a} e^{- x y} d y) d x = \int_{0}^{\infty} (\frac{e^{- x} - e^{- a x}}{x}) d x = I (a)$

Conclusion: The equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$ is mathematically proved.

To determine

(d)

By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

Expert Solution

Answer to Problem 53E

Solution: By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is derived.

Explanation of Solution

Given: $I (a) = \int_{0}^{T} \int_{1}^{a} e^{- x y} d y d x$ , a>0

Calculation:

Step 1: By the definition of the improper integral,

$I (a) = \lim_{T \to \infty} \int_{0}^{T} \int_{1}^{a} e^{- x y} d y d x$

Step 2: We compute the double integral. Using Fubini's Theorem we may compute the iterated integral using the reversed order of integration. That is,

$\begin{array}{l} \int_{0}^{T} \int_{1}^{a} e^{- x y} d y d x = \int_{1}^{a} \int_{0}^{T} e^{- x y} d x d y = \int_{1}^{a} (\int_{0}^{T} e^{- x y} d x) d y = \int_{1}^{a} ({- \frac{1}{y} e^{- x y} |}_{x = 0}^{T}) d y \\ = \int_{1}^{a} (- \frac{1}{y} (e^{- T y} - e^{- 0. y})) d y = \int_{1}^{a} \frac{1 - e^{- T y}}{y} d y = \int_{1}^{a} \frac{d y}{y} - \int_{1}^{a} \frac{e^{- T y}}{y} d y \\ = {\ln y |}_{1}^{a} - \int_{1}^{a} \frac{e^{- T y}}{y} d y = \ln a - \ln 1 - \int_{1}^{a} \frac{e^{- T y}}{y} d y = \ln a - \int_{1}^{a} \frac{e^{- T y}}{y} d y \end{array}$

Combining with Step 1, we get,

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

Conclusion: By interchanging the order of integration, the equation $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is mathematically proved.

To determine

(e)

The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem.

Expert Solution

Answer to Problem 53E

Solution: The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem is derived.

Explanation of Solution

Given: $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ , a>0

Calculation:

Step 1: We consider the following possible cases:

Case 1: $a \geq 1$ then in the interval of integration $y \geq 1$ . As $T \to \infty$ , we may assume that T > 0

Thus,

$\frac{e^{- T y}}{y} \leq \frac{e^{- T .1}}{1} = e^{- T}$

Hence,

$0 \leq \int_{1}^{a} \frac{e^{- T y}}{y} d y \leq \int_{1}^{a} e^{- T} d y = e^{- T} (a - 1)$

By the limit $\lim_{T \to \infty} e^{- T} (a - 1) = 0$ and the Squeeze Theorem, we conclude that,

$\lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y = 0$

Case 2: $0 < a < 1.$

Then,

$\int_{1}^{a} \frac{e^{- T y}}{y} d y = - \int_{a}^{1} \frac{e^{- T y}}{y} d y$

and in the interval of integration $a \leq y \leq 1$ , therefore

$\frac{e^{- T y}}{y} \leq \frac{e^{- T a}}{a}$ (the function $\frac{e^{- T y}}{y}$ is decreasing).

Hence,

$0 \leq \int_{a}^{1} \frac{e^{- T y}}{y} d y \leq \int_{a}^{1} \frac{e^{- T a}}{a} d y = \frac{1 - a}{a} e^{- T a}$

By the limit $\lim_{T \to \infty} \frac{1 - a}{a} e^{- T a} = 0$ and the Squeeze Theorem, we conclude that

$\lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} = - \lim_{T \to \infty} \int_{a}^{1} \frac{e^{- T y}}{y} = 0$

We thus showed that for all a > 0, $\lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} = 0$

Conclusion: The statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem is mathematically proved.

Answer 6

Chapter 15.1, Problem 53E

To determine

(a)

The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$

Expert Solution

Answer to Problem 53E

Solution: The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$ is derived.

Explanation of Solution

Explanation:

Given: $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ for a>0

Calculation:

Step 1: The function $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , $f (0) = a - 1$ is continuous if $\lim_{x \to 0} f (x) = f (0) = a - 1$ .

Step 2: We verify this limit using L'Hopital Rule:

$\lim_{x \to 0} \frac{e^{- x} - e^{- a x}}{x} = \lim_{x \to 0} \frac{- e^{- x} + a e^{- a x}}{1} = - 1 + a = a - 1$

Therefore, f is continuous.

Conclusion: The statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$ is mathematically proved.

Answer 7

Chapter 15.1, Problem 53E

To determine

(a)

The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$

Answer 8

Expert Solution

Answer to Problem 53E

Solution: The mathematical proof of the statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$ is derived.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Explanation:

Given: $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ for a>0

Calculation:

Step 1: The function $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , $f (0) = a - 1$ is continuous if $\lim_{x \to 0} f (x) = f (0) = a - 1$ .

Step 2: We verify this limit using L'Hopital Rule:

$\lim_{x \to 0} \frac{e^{- x} - e^{- a x}}{x} = \lim_{x \to 0} \frac{- e^{- x} + a e^{- a x}}{1} = - 1 + a = a - 1$

Therefore, f is continuous.

Conclusion: The statement stating that using L'Hopital Rule, $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , though not defined at x = 0, can be made continuous by assigning the value $f (0) = a - 1$ is mathematically proved.

Answer 13

To determine

(b)

The mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem

Expert Solution

Answer to Problem 53E

Solution: Both the mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem are derived.

Explanation of Solution

Given: $x > 1$ , $I (a) = \int_{0}^{\infty} \frac{e^{- x} - e^{- a x}}{x} d x$ , $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , a > 0

Calculation:

Step 1: We now show that the following integral converges:

$I (a) = \int_{0}^{\infty} \frac{e^{- x} - e^{- a x}}{x} d x$ (a>0)

Since, $e^{- x} - e^{- a x} < e^{- x} + e^{- a x}$ then $\frac{e^{- x} - e^{- a x}}{x} < \frac{e^{- x} + e^{- a x}}{x}$ for x > 0

If x > 1 we have,

$\frac{e^{- x} - e^{- a x}}{x} < \frac{e^{- x} + e^{- a x}}{x} < e^{- x} + e^{- a x}$

That is for x > 1

$f (x) < e^{- x} + e^{- a x}$

Step 2: Also, since $e^{- a x} - e^{- x} < e^{- a x} + e^{- x}$ we have for x > 1

$\frac{e^{- a x} - e^{- x}}{x} < \frac{e^{- a x} - e^{- x}}{x} < e^{- a x} + e^{- x}$

Thus, we get

$- f (x) < e^{- x} + e^{- a x}$

Step 3: Hence, from Step 1 and Step 2, we get

$0 \leq | f (x) | \leq e^{- x} + e^{- a x}$

Step 4: We now show that the integral of the right hand side converges:

$\begin{array}{l} \int_{0}^{\infty} (e^{- x} + e^{- a x}) d x = \lim_{R \to \infty} \int_{0}^{R} (e^{- x} + e^{- a x}) d x = \lim_{R \to \infty} (- e^{- x} - {\frac{e^{- a x}}{a} |}_{x = 0}^{R}) \\ = \lim_{R \to \infty} (e^{- R} - \frac{e^{- a R}}{a} + e^{0} + \frac{e^{0}}{a}) = \lim_{R \to \infty} (e^{- R} - \frac{e^{- a R}}{a} + 1 + \frac{1}{a}) = 1 + \frac{1}{a} \end{array}$

Since the integral converges, we conclude from Step 3 and the Comparison Test for Improper Integral that

$\int_{0}^{\infty} \frac{(e^{- x} + e^{- a x})}{x} d x$ also converges for a > 0.

Conclusion: Both the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the statement stating that I(a) converges by applying the Comparison Theorem are mathematically proved.

Answer 14

To determine

(b)

The mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem

Answer 15

Expert Solution

Answer to Problem 53E

Solution: Both the mathematical proof of the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem are derived.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given: $x > 1$ , $I (a) = \int_{0}^{\infty} \frac{e^{- x} - e^{- a x}}{x} d x$ , $f (x) = \frac{e^{- x} - e^{- a x}}{x}$ , a > 0

Calculation:

Step 1: We now show that the following integral converges:

$I (a) = \int_{0}^{\infty} \frac{e^{- x} - e^{- a x}}{x} d x$ (a>0)

Since, $e^{- x} - e^{- a x} < e^{- x} + e^{- a x}$ then $\frac{e^{- x} - e^{- a x}}{x} < \frac{e^{- x} + e^{- a x}}{x}$ for x > 0

If x > 1 we have,

$\frac{e^{- x} - e^{- a x}}{x} < \frac{e^{- x} + e^{- a x}}{x} < e^{- x} + e^{- a x}$

That is for x > 1

$f (x) < e^{- x} + e^{- a x}$

Step 2: Also, since $e^{- a x} - e^{- x} < e^{- a x} + e^{- x}$ we have for x > 1

$\frac{e^{- a x} - e^{- x}}{x} < \frac{e^{- a x} - e^{- x}}{x} < e^{- a x} + e^{- x}$

Thus, we get

$- f (x) < e^{- x} + e^{- a x}$

Step 3: Hence, from Step 1 and Step 2, we get

$0 \leq | f (x) | \leq e^{- x} + e^{- a x}$

Step 4: We now show that the integral of the right hand side converges:

$\begin{array}{l} \int_{0}^{\infty} (e^{- x} + e^{- a x}) d x = \lim_{R \to \infty} \int_{0}^{R} (e^{- x} + e^{- a x}) d x = \lim_{R \to \infty} (- e^{- x} - {\frac{e^{- a x}}{a} |}_{x = 0}^{R}) \\ = \lim_{R \to \infty} (e^{- R} - \frac{e^{- a R}}{a} + e^{0} + \frac{e^{0}}{a}) = \lim_{R \to \infty} (e^{- R} - \frac{e^{- a R}}{a} + 1 + \frac{1}{a}) = 1 + \frac{1}{a} \end{array}$

Since the integral converges, we conclude from Step 3 and the Comparison Test for Improper Integral that

$\int_{0}^{\infty} \frac{(e^{- x} + e^{- a x})}{x} d x$ also converges for a > 0.

Conclusion: Both the statement stating that $| f (x) | \leq e^{- x} + e^{- a x}$ for $x > 1$ using triangle inequality and the statement stating that I(a) converges by applying the Comparison Theorem are mathematically proved.

Answer 20

To determine

(c)

The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$

Expert Solution

Answer to Problem 53E

Solution: The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$ is derived

Explanation of Solution

Given: $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$ , a>0

Calculation:

Step 1: We compute the inner integral with respect to y:

$\int_{1}^{a} e^{- x y} d y = - \frac{1}{x} {e^{- x y} |}_{y = 1}^{a} = - \frac{1}{x} (e^{- x a} - e^{- x .1}) = \frac{e^{- x} - e^{- a x}}{x}$

Step 2: Hence,

$I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x = \int_{0}^{\infty} (\int_{1}^{a} e^{- x y} d y) d x = \int_{0}^{\infty} (\frac{e^{- x} - e^{- a x}}{x}) d x = I (a)$

Conclusion: The equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$ is mathematically proved.

Answer 21

To determine

(c)

The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$

Answer 22

Expert Solution

Answer to Problem 53E

Solution: The mathematical proof of the equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$ is derived

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given: $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$ , a>0

Calculation:

Step 1: We compute the inner integral with respect to y:

$\int_{1}^{a} e^{- x y} d y = - \frac{1}{x} {e^{- x y} |}_{y = 1}^{a} = - \frac{1}{x} (e^{- x a} - e^{- x .1}) = \frac{e^{- x} - e^{- a x}}{x}$

Step 2: Hence,

$I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x = \int_{0}^{\infty} (\int_{1}^{a} e^{- x y} d y) d x = \int_{0}^{\infty} (\frac{e^{- x} - e^{- a x}}{x}) d x = I (a)$

Conclusion: The equation $I (a) = \int_{0}^{\infty} \int_{1}^{a} e^{- x y} d y d x$ is mathematically proved.

Answer 27

To determine

(d)

By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

Expert Solution

Answer to Problem 53E

Solution: By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is derived.

Explanation of Solution

Given: $I (a) = \int_{0}^{T} \int_{1}^{a} e^{- x y} d y d x$ , a>0

Calculation:

Step 1: By the definition of the improper integral,

$I (a) = \lim_{T \to \infty} \int_{0}^{T} \int_{1}^{a} e^{- x y} d y d x$

Step 2: We compute the double integral. Using Fubini's Theorem we may compute the iterated integral using the reversed order of integration. That is,

$\begin{array}{l} \int_{0}^{T} \int_{1}^{a} e^{- x y} d y d x = \int_{1}^{a} \int_{0}^{T} e^{- x y} d x d y = \int_{1}^{a} (\int_{0}^{T} e^{- x y} d x) d y = \int_{1}^{a} ({- \frac{1}{y} e^{- x y} |}_{x = 0}^{T}) d y \\ = \int_{1}^{a} (- \frac{1}{y} (e^{- T y} - e^{- 0. y})) d y = \int_{1}^{a} \frac{1 - e^{- T y}}{y} d y = \int_{1}^{a} \frac{d y}{y} - \int_{1}^{a} \frac{e^{- T y}}{y} d y \\ = {\ln y |}_{1}^{a} - \int_{1}^{a} \frac{e^{- T y}}{y} d y = \ln a - \ln 1 - \int_{1}^{a} \frac{e^{- T y}}{y} d y = \ln a - \int_{1}^{a} \frac{e^{- T y}}{y} d y \end{array}$

Combining with Step 1, we get,

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

Conclusion: By interchanging the order of integration, the equation $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is mathematically proved.

Answer 28

To determine

(d)

By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

Answer 29

Expert Solution

Answer to Problem 53E

Solution: By interchanging the order of integration, the mathematical proof of the equation

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is derived.

Answer 30

Expert Solution

Answer 31

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given: $I (a) = \int_{0}^{T} \int_{1}^{a} e^{- x y} d y d x$ , a>0

Calculation:

Step 1: By the definition of the improper integral,

$I (a) = \lim_{T \to \infty} \int_{0}^{T} \int_{1}^{a} e^{- x y} d y d x$

Step 2: We compute the double integral. Using Fubini's Theorem we may compute the iterated integral using the reversed order of integration. That is,

$\begin{array}{l} \int_{0}^{T} \int_{1}^{a} e^{- x y} d y d x = \int_{1}^{a} \int_{0}^{T} e^{- x y} d x d y = \int_{1}^{a} (\int_{0}^{T} e^{- x y} d x) d y = \int_{1}^{a} ({- \frac{1}{y} e^{- x y} |}_{x = 0}^{T}) d y \\ = \int_{1}^{a} (- \frac{1}{y} (e^{- T y} - e^{- 0. y})) d y = \int_{1}^{a} \frac{1 - e^{- T y}}{y} d y = \int_{1}^{a} \frac{d y}{y} - \int_{1}^{a} \frac{e^{- T y}}{y} d y \\ = {\ln y |}_{1}^{a} - \int_{1}^{a} \frac{e^{- T y}}{y} d y = \ln a - \ln 1 - \int_{1}^{a} \frac{e^{- T y}}{y} d y = \ln a - \int_{1}^{a} \frac{e^{- T y}}{y} d y \end{array}$

Combining with Step 1, we get,

$I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$

Conclusion: By interchanging the order of integration, the equation $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is mathematically proved.

Answer 34

To determine

(e)

The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem.

Expert Solution

Answer to Problem 53E

Solution: The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem is derived.

Explanation of Solution

Given: $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ , a>0

Calculation:

Step 1: We consider the following possible cases:

Case 1: $a \geq 1$ then in the interval of integration $y \geq 1$ . As $T \to \infty$ , we may assume that T > 0

Thus,

$\frac{e^{- T y}}{y} \leq \frac{e^{- T .1}}{1} = e^{- T}$

Hence,

$0 \leq \int_{1}^{a} \frac{e^{- T y}}{y} d y \leq \int_{1}^{a} e^{- T} d y = e^{- T} (a - 1)$

By the limit $\lim_{T \to \infty} e^{- T} (a - 1) = 0$ and the Squeeze Theorem, we conclude that,

$\lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y = 0$

Case 2: $0 < a < 1.$

Then,

$\int_{1}^{a} \frac{e^{- T y}}{y} d y = - \int_{a}^{1} \frac{e^{- T y}}{y} d y$

and in the interval of integration $a \leq y \leq 1$ , therefore

$\frac{e^{- T y}}{y} \leq \frac{e^{- T a}}{a}$ (the function $\frac{e^{- T y}}{y}$ is decreasing).

Hence,

$0 \leq \int_{a}^{1} \frac{e^{- T y}}{y} d y \leq \int_{a}^{1} \frac{e^{- T a}}{a} d y = \frac{1 - a}{a} e^{- T a}$

By the limit $\lim_{T \to \infty} \frac{1 - a}{a} e^{- T a} = 0$ and the Squeeze Theorem, we conclude that

$\lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} = - \lim_{T \to \infty} \int_{a}^{1} \frac{e^{- T y}}{y} = 0$

We thus showed that for all a > 0, $\lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} = 0$

Conclusion: The statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem is mathematically proved.

Answer 35

To determine

(e)

The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem.

Answer 36

Expert Solution

Answer to Problem 53E

Solution: The mathematical proof of the statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem is derived.

Answer 37

Expert Solution

Answer 38

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Given: $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ , a>0

Calculation:

Step 1: We consider the following possible cases:

Case 1: $a \geq 1$ then in the interval of integration $y \geq 1$ . As $T \to \infty$ , we may assume that T > 0

Thus,

$\frac{e^{- T y}}{y} \leq \frac{e^{- T .1}}{1} = e^{- T}$

Hence,

$0 \leq \int_{1}^{a} \frac{e^{- T y}}{y} d y \leq \int_{1}^{a} e^{- T} d y = e^{- T} (a - 1)$

By the limit $\lim_{T \to \infty} e^{- T} (a - 1) = 0$ and the Squeeze Theorem, we conclude that,

$\lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y = 0$

Case 2: $0 < a < 1.$

Then,

$\int_{1}^{a} \frac{e^{- T y}}{y} d y = - \int_{a}^{1} \frac{e^{- T y}}{y} d y$

and in the interval of integration $a \leq y \leq 1$ , therefore

$\frac{e^{- T y}}{y} \leq \frac{e^{- T a}}{a}$ (the function $\frac{e^{- T y}}{y}$ is decreasing).

Hence,

$0 \leq \int_{a}^{1} \frac{e^{- T y}}{y} d y \leq \int_{a}^{1} \frac{e^{- T a}}{a} d y = \frac{1 - a}{a} e^{- T a}$

By the limit $\lim_{T \to \infty} \frac{1 - a}{a} e^{- T a} = 0$ and the Squeeze Theorem, we conclude that

$\lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} = - \lim_{T \to \infty} \int_{a}^{1} \frac{e^{- T y}}{y} = 0$

We thus showed that for all a > 0, $\lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} = 0$

Conclusion: The statement stating the limit in $I (a) = \ln a - \lim_{T \to \infty} \int_{1}^{a} \frac{e^{- T y}}{y} d y$ is zero by using the Comparison Theorem is mathematically proved.

Answer 41

Ch. 15.1 - Prob. 1PQ Ch. 15.1 - Prob. 2PQ Ch. 15.1 - Prob. 3PQ Ch. 15.1 - Prob. 4PQ Ch. 15.1 - Prob. 5PQ Ch. 15.1 - Prob. 6PQ Ch. 15.1 - Prob. 1E Ch. 15.1 - Prob. 2E Ch. 15.1 - Prob. 3E Ch. 15.1 - Prob. 4E

(a) The mathematical proof of the statement stating that using L'Hopital Rule, f ( x ) = e − x − e − a x x , though not defined at x = 0, can be made continuous by assigning the value f ( 0 ) = a − 1